What is the sum from n=1 to infinity of (3^n)/(12^n) ?

The sum from n=1 to infinity of (3^n)/(12^n) is 1/3

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sum from n=1 to infinity of (3^n)/(12^n)

n = 1 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

\frac{1}{3}

Solution steps

n = 1 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

Simplify

\frac{3 ^{n}}{1 2 ^{n}} : \frac{1}{4 ^{n}}

= n = 1 \sum \infty \frac{1}{4 ^{n}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty \frac{1}{4 ^{n}} - \frac{1}{4 ^{0}}

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= \frac{4}{3} - \frac{1}{4 ^{0}}

Simplify

\frac{4}{3} - \frac{1}{4 ^{0}} : \frac{1}{3}

= \frac{1}{3}

n = 1 \sum \infty \frac{4}{7 n - 1}

n = 9 \sum \infty \frac{4}{8 ^{n}}

n = 0 \sum \infty \frac{n ^{2}}{1}

k = 1 \sum \infty 4^{- k}

k = 0 \sum \infty \frac{1}{900 + 3 k}

What is the sum from n=1 to infinity of (3^n)/(12^n) ?
The sum from n=1 to infinity of (3^n)/(12^n) is 1/3